A cube of side a has point charges +Q located | vedclass

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(B) We can replace the $-Q$ charge at the origin by adding $+Q$ and $-2Q$ at that point. Now,due to $+Q$ charges at all $8$ corners of the cube,the el

Vedclass · Accepted Answer

$\frac{-2 Q}{3 \sqrt{3} \pi \varepsilon_{0} a^{2}}(\hat{x}+\hat{y}+\hat{z})$

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(B) We can replace the $-Q$ charge at the origin by adding $+Q$ and $-2Q$ at that point. Now,due to $+Q$ charges at all $8$ corners of the cube,the electric field at the center of the cube is zero by symmetry. Thus,the net electric field at the center is only due to the $-2Q$ charge placed at the origin. The position vector of the center of the cube relative to the origin is $\vec{r} = \frac{a}{2}\hat{x} + \frac{a}{2}\hat{y} + \frac{a}{2}\hat{z} = \frac{a}{2}(\hat{x} + \hat{y} + \hat{z})$. The distance from the origin to the center is $r = \sqrt{(\frac{a}{2})^2 + (\frac{a}{2})^2 + (\frac{a}{2})^2} = \frac{a\sqrt{3}}{2}$. The electric field $\vec{E}$ due to a point charge $q$ is given by $\vec{E} = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^3} \vec{r}$. Substituting $q = -2Q$ and the vector $\vec{r}$,we get: $\vec{E} = \frac{1}{4\pi\varepsilon_0} \frac{-2Q}{(\frac{a\sqrt{3}}{2})^3} \cdot \frac{a}{2}(\hat{x} + \hat{y} + \hat{z})$ $\vec{E} = \frac{-2Q}{4\pi\varepsilon_0} \cdot \frac{8}{3\sqrt{3}a^3} \cdot \frac{a}{2}(\hat{x} + \hat{y} + \hat{z})$ $\vec{E} = \frac{-2Q}{3\sqrt{3}\pi\varepsilon_0 a^2}(\hat{x} + \hat{y} + \hat{z})$

$A$ cube of side $a$ has point charges $+Q$ located at each of its vertices except at the origin where the charge is $-Q$. The electric field at the centre of the cube is

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